ENTROPY AND CALCULATIONS

This is defined as a measure of the degree of disorderliness or randomness of a system. Any spontaneous change in an isolated system must be accompanied by an increase in entropy. Systems having perfect order, i.e. pure crystalline compounds and elements at zero Kelvin have zero entropy. Entropy increases in going from solid to liquid and to gas. Like enthalpy (H), entropy (S) depends on the initial and final states of the system undergoing change (i.e. entropy is also a state function).             

          

Thus;   ∆S = ƐnS_(product)  -  ƐmS_{(reactant)........}1

Where n and m are the coefficients of the substances in the product(s) and reactant(s) respectively; Ɛ represents summation.

WORKED EXAMPLES

Question1: consider the reactions given below, will you expect entropy to increase of decrease? Explain your answer.

i.                     NH_3(g) + HCL_(g) ------------ NH₄CL_(s)

ii.                   CO_(g) + Br_2(g) -------------- COBr_2(g)         

      

SOLUTION:

I.                    NH_3(g) + HCL_(g) ------------ NH₄CL_(s)

For the above chemical reaction (heterogeneous system); both reactant are in gaseous state, and hence there is more disorderliness (more entropy) compared with the product which is in a solid state, with less disorderliness. Therefore using equation 1 above, the entropy for the above chemical equation would be negative (i.e. decrease in entropy as the reaction proceeds to the formation of the product).

ii       CO_(g) + Br_2(g) -------------- COBr_2(g)          

For the above chemical reaction, both reactants and product are in gaseous phase (homogeneous system). Using the same equation 1 above, the entropy change for the above reaction would be negative (i.e. decrease), since there are more gaseous volume of reactants than the products.

ENTROPY CHANGE FOR TRANSITION PHASE

Consider the phase transition:

                                        CO_2(S)                  CO_2(g)

For the above transition (i.e. sublimation), solid CO₂ passes directly into gaseous CO₂. All phase transition are always in equilibrium with each other; hence the free energy change is zero, i.e.

For phase transition;   ∆G = 0  --------------------- 2

Enthalpy change and entropy change are related to free energy by the following expression already given by the equation below:

                                                              ∆G = ∆H - T∆S

                                                        T_t∆S_t = ∆H_t (if ∆G = 0).

For phase transition;

                                ∆G = 0 = ∆H_t - T_t∆S_t

Therefore, ∆S_t =  ∆H_t/T_t

Where,

∆S_t = entropy change accompanying the reaction

∆H_t = heat change accompanying the transition

T_t = temperature of transition

WORKED EXAMPLE

Question2: the heat of vapourization, ∆H_vap of carbon tetrachloride, CCL₄ at 25^oc is 43kj/k/mol. If 1 mole of liquid CCL₄ at 25^oC has an entropy of 214j/k. what is the entropy of 1 mole of the vapour in equilibrium with the liquid at this temperature.

SOLUTION:

∆S_vap. = ∆H_vap/T

            =  43 X 10³ J/mol/298k

∆S_vap. = 144 J/K/mol.

Therefore, 1 mole of CCL₄ increases in entropy by 144 J/k when it vapourizes. The entropy of 1 mole of vapour in equilibrium with the liquid = 1 mole of liquid + 144J/K

                                                                      = (214 + 144) J/K/mol

                                                                       = 358 J/K/mol.